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Re: How many times will the digit 7 be written? [#permalink]
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10 Oct 2012, 04:56
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MOST EFFICIENT METHOD USING COMBINATORIXAs Bunuel said, think of each number as XXX ie 001 to 999 Numbers with one 7. Either 7XX, X7X or XX7, so 3 x 1C9 x 1C9 = 3x9x9 = 243 Numbers with two 7's (nb question asks how many sevens are displayed, so this answer must be doubled) Either 77X, 7X7 or X77, so 3x 1C9 = 27 (or 54 once you double it) Numbers with three 7's Only 1 (777). Triple it as there are three 7,s = 3 Add all together: 243+54+3 = 300 Answer => D
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Re: counting [#permalink]
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29 Oct 2012, 12:16
tt11234 wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
110 111 271 300 304
any easy way to do this question? here is how i did it... let xyz be a 3digit#, first, let Z=7, we have 10 options ( 09)for x and 10 options for y, then we get 10 x 10 =100, same logic is applied when y =7 and x = 7, we get 300 times. is it correct? started using the same method, but got stuck in the middle of the road. Bunuel, could you please clarify if slot method can be applied to this problem. Thanks in advance.



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Re: counting [#permalink]
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29 Oct 2012, 12:33
EV wrote: tt11234 wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
110 111 271 300 304
any easy way to do this question? here is how i did it... let xyz be a 3digit#, first, let Z=7, we have 10 options ( 09)for x and 10 options for y, then we get 10 x 10 =100, same logic is applied when y =7 and x = 7, we get 300 times. is it correct? started using the same method, but got stuck in the middle of the road. Bunuel, could you please clarify if slot method can be applied to this problem. Thanks in advance. Check here: howmanytimeswillthedigit7bewritten99914.html#p781456Another approach: howmanytimeswillthedigit7bewritten99914.html#p770507
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Re: How many times will the digit 7 be written? [#permalink]
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08 Jul 2013, 00:10



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Re: How many times will the digit 7 be written? [#permalink]
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14 Jul 2014, 16:54
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Alternate solution:
Using probability of occurrence = desired outcomes / total outcomes, desired outcomes = probability x total outcomes.
total outcomes = 1000.
probability of occurrence  to simplify this, the question statement could be interpreted as "what is the probability that the digit 7 is picked at least once when 3 digits are chosen at random?"
probability of at least one 7 = 7 in first digit OR 7 in second digit OR 7 in third digit = 1/10 + 1/10 + 1/10 = 3/10
desired outcomes = probability x total outcomes = 3/10 x 1000 = 300.



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Re: How many times will the digit 7 be written? [#permalink]
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17 Jul 2014, 21:23
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The first time I did the question, my answer was C : 271 since I thought the question is "how many numbers from 0  1000, that contain the 7 digit". So I counted 77, 707,717,....,777,...797, 771, 772, 773,..., or 779 as one. That may be the reason why 23% of people who answered the question decided to pick answer C. That is really interesting question
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Re: How many times will the digit 7 be written? [#permalink]
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It is very interesting question, thanks for the all the suggestion but Banuel is relly a rockstar.. how to promote an app



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Re: How many times will the digit 7 be written? [#permalink]
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18 Jul 2014, 13:23
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vad3tha wrote: The first time I did the question, my answer was C : 271 since I thought the question is "how many numbers from 0  1000, that contain the 7 digit". So I counted 77, 707,717,....,777,...797, 771, 772, 773,..., or 779 as one. That may be the reason why 23% of people who answered the question decided to pick answer C. That is really interesting question Yes, I made the same mistake.



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Re: How many times will the digit 7 be written? [#permalink]
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11 Dec 2014, 23:30
Bunuel wrote: nonameee wrote: Bunuel, can you please explain the logic here? I don't understand it. Quote: Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times... Not sure what can I add to this... It just means that out of 3000 digits we used to write down first 1000 numbers (000, 001, 002, 003, ..., 999), each digit from 0 to 9 is used equal number of times, how else? Thus we used each digit 3000/10=300 times. HI Bunuel, I also applied same approach. I got 19 7's per 100. here you have mentioned 20 so why are we counting 77 2 times? Please clarify. Thanks



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Re: How many times will the digit 7 be written? [#permalink]
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How many times will the digit 7 be written? [#permalink]
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27 Mar 2015, 23:39
Another approach: Think of the digits in the form: abcd where a can only be 0 (since 1000 doesn't contain any 7s) and b,c, and d can be any digit from 09.
Written Once Ones Place=(1*9*9*1)*1=81 Tens Place= (1*9*1*9)*1=81 Hundreds Place= (1*1*9*9)*1=81
Written Twice Ones & Tens=(1*9*1*1)*2=18 Ones & Hundreds=(1*1*9*1)*2=18 Tens & Hundreds=(1*1*1*9)*2=18
Written Three Times (0777) Ones & Tens & Hundreds=(1*1*1*1)*3
81*3+18*3+3=300



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Re: How many times will the digit 7 be written? [#permalink]
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30 Mar 2015, 23:26
Let's take numbers from 700 till 799: We have 100 numbers containing 7 as a hundredth digit We have 10 additional 7 if we take the second digit, and 10 more for the units digit Therefore we have 100+10+10=120 7s.
Let's take numbers from 800 till 899: We have 10 numbers with 7 if we take the second digit, and 10 more for the units digit for a total of 20. The same can be applied for the rest of the hundredth which are 8: 000, 100, 200, 300, 400, 500, 600, 900 So in total we have 9*20+120=300



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How many times will the digit 7 be written? [#permalink]
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03 Jul 2015, 08:34
Bunuel wrote: seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? (A) 110 (B) 111 (C) 271 (D) 300 (E) 304
Can someone explain a simpler approach to this problem? Many approaches are possible. For example: Consider numbers from 0 to 999 written as follows: 1. 000 2. 001 3. 002 4. 003 ... ... ... 1000. 999 We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times. Answer: D. I get how the answer is 300 and that you intentionally start with 000 instead of 001... but I disagree that no digit has preference over another (since the number range is 11000 and not 000999, so the former's number of digits matters the most, esp. if they ask how many 1s digits there are). The 1 digit is used 301 times, not 300 (you have to include the 1 in 1,000). The 0 digit does show up 300 times, but it assumes that we use that method. Even if we assume the method, it depends on which segment of the numbers you look at, esp. if the GMAT asks a variation of this question that shows this difference as outlined below (in which case, you should start with 001 and not 000 so there are 3 less 0s in the first 99 numbers, compared with the three extra 0s in the very last number). So, for the 0s, you can't simply say that 001099 has the same number of 0s as the 1s in the range 100199. That gives us a total of 3001 digits. Yet, even this total is misleading... if the GMAT asks how many times the number 0 shows up in any of the digits from 11000, you wouldn't start with "001". Rather, you'd start with "1". Then, the number of 0's in total is 192, and consequently, the total number of digits is 2893. Only the digits 29 show up the same number of times between 11000. If you want to be a bit more conservative about it, you should only use the method when the same number of digits are used for the lower and upper limits of the range (e.g. 1099; 100999), then use the method again for the single digits, the tens digits, and the thousand digits, and add it all up. The "1" digit shows up this many times per given interval below: 001  099 x20 100  199 x120 = 100, (looking at just the units/tens digits) 01, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91, all the 1s in the hundred digit is 1001s 200  299 x20300  399 x20400  499 x20500  599 x20600  699 x20700  799 x20800  899 x20900  999 x201000 x1Total = 301The "0" digit shows up this many times per given interval below, though you should not use this method for the aforementioned reason: 001  099 x117 = (looking at just the units/tens digits) 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40, 50, 60, 70, 80, 90, all the 0s in the hundred digit is 990s (it's 99 since we can't include 000 and there are only 99 numbers  if we did include 000 it would be 1000s in the hundred digit + the two other 0s in the number 1000) 100  199 x20 200  299 x20300  399 x20400  499 x20500  599 x20600  699 x20700  799 x20800  899 x20900  999 x201000 x3Total = 300Conversely, if they ask how many times the number 0 shows up in each digit from 11000, it would be: 1  99 x9 = 10, 20, 30, 40, 50, 60, 70, 80, 90 100  199 x20 = 100 (two 0s), 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190 200  299 x20300  399 x20400  499 x20500  599 x20600  699 x20700  799 x20800  899 x20900  999 x201000 x3Total = 192Or simply: \([\frac{(10^n)}{10}]*n  111*(1  \frac{x}{x})\), where n is the total number of complete set of digits from 100 onward (so, up to 999 counts as 3, but up to 1,000 doesn't count as 4 as it would need to be 9,999 to count as 4), and x is the number that you want to find repeating (e.g. 09). I made up this equation but I think it works... you just need to ignore the undefined part and make it equal to 0 if you were to select 0... So, if you try it for 1  99,999,999, and see how many times the number 4 appears, then it's: \([\frac{(10^8)}{10}]*8  111*(1  \frac{5}{5}) = 8*10^7\) times the number 4 shows up between 1 and 99,999,999 inclusive.



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Re: How many times will the digit 7 be written? [#permalink]
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17 Oct 2015, 03:10
Bunuel wrote: seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? (A) 110 (B) 111 (C) 271 (D) 300 (E) 304
Can someone explain a simpler approach to this problem? Many approaches are possible. For example: Consider numbers from 0 to 999 written as follows: 1. 000 2. 001 3. 002 4. 003 ... ... ... 1000. 999 We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times. Answer: D. my approach to this question was from permutation view I took numbers from 0 to 999 (instead of from 1 to 1000), for easiness we are taking here 0 to be as "000" , i.e. all digits as three digits, we can not neglect 0's in the beginning now we have three places, and every digit from 0 to 9 has equal chance to come on every place so, out of total 1000 numbers 7 comes in hundreds place 100 times out of total 1000 numbers 7 comes in tens place for 100 times and out of total 1000 numbers 7 comes in once place for 100 times so, we get total of 300, Answer D Kudos, if you like the post helpful



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Re: How many times will the digit 7 be written? [#permalink]
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31 Jan 2016, 14:36
Bunuel wrote: seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? (A) 110 (B) 111 (C) 271 (D) 300 (E) 304
Can someone explain a simpler approach to this problem? Many approaches are possible. For example: Consider numbers from 0 to 999 written as follows: 1. 000 2. 001 3. 002 4. 003 ... ... ... 1000. 999 We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times. Answer: D. ingenious solution!



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How many times will the digit 7 be written? [#permalink]
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26 Nov 2016, 07:25
I think this is easiest to think with the counting formula/boxes.
How many times 7 appears b/w 11000. Subtract 1 from 1000 to get 999. Now you are working with a 3 digit number.
Case 1: _ _ _ First digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (7XX) You have 10 options for the second digit. You have 10 options for the third digit. Case 1 total: 1x10x10 = 100
Case 2: _ _ _ Second digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (X7X) You have 10 options for the first digit. You have 10 options for the third digit. Case 2 total: 10x1x10 = 100
Case 3: _ _ _ Last digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (XX7) You have 10 options for the first digit. You have 10 options for the second digit. Case 3 total: 10x10x1 = 100
Add all case totals together: Case 1(100) + Case 2(100) + Case 3(100) = 300 final answer.
Is this a correct way to think about it?



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Re: How many times will the digit 7 be written? [#permalink]
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01 Jun 2017, 02:07
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 304 We can solve using permutation (order does matter in numbers): Possibilities include numbers of 3 digits (from 1 to 999) Either one of the 3 digits has to be 7 (Permutation of a 1 P 3, no repetition) = 3!/(31)!=3 The 2 remaining digits can be any other number from 0 to 9 (Permutation of 2 numbers from 10, with repetition = 10^2) Total = 3 x 10^2 = 300



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Re: How many times will the digit 7 be written? [#permalink]
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22 Jan 2018, 15:08
Hi All, There are a couple of different way to conceive of this question, and each has its own 'organization' to it, so you should try to think in whatever terms are easiest for you. To start, it shouldn't be hard to figure out how many 3DIGIT numbers will START with 7... 700 to 799 inclusive... so that's 100 appearances of a 7 right there. Next, you might find it easiest to think about the UNITs DIGIT. Consider the following pattern... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 etc Notice how 1 out of every 10 numbers has a unit's digit of 7? That pattern repeats overandover. Since we have 1000 total consecutive integers in our list, 1/10 of them will have a unit's digit of 7... (1000)(1/10) = 100 appears of a 7... Now, think about what you've seen so far... a group of 100 and another group of 100. I wonder what will happen when we deal with the TENS DIGITS... 10 20 30 40 50 60 70 80 90 100 __ 71 72 73 74 75 76 7_ 78 79 80 In the first 100 integers, there are 10 additional 7s in the TENS 'spot' (I removed 70 and the units digit 7 from 77 since I already counted those). There are 10 sets of 100 to consider, so there are (10)(10) = 100 additional 7s... Total = 100 + 100 + 100 = 300 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: How many times will the digit 7 be written? [#permalink]
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24 Jan 2018, 05:45
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 304 From 1 to 100  7 will be written 20 times. So, from 1 to 999, the last 2 Digits containing 7 will 10*20 = 200. Also, from 700 to 799.. we have hundredth digit written 100 times .. So, total will be 200 + 100 = 300.
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Re: How many times will the digit 7 be written? [#permalink]
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17 Feb 2018, 12:52
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 304 I solved it in 19seconds through the following approach:  In 10 numbers, 7 is used 1 time.  In 100 numbers, 7 is used 1*10 times + the 10 times from 70, so 20 times.  In 1000 numbers, 7 is used 20*10 + the 100 times from 700, so 300 times.  In 10000 numbers, 7 is used 300*10 + the 1000 times from 7000, so 4000 times.  Etc.




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